Kinematics Previous Year Questions — JEE Main Physics
38 past-year Kinematics questions from JEE Main (Physics) — page 8.
Q44 (2026)
The velocity (v) - Distance (x) graph is shown in figure. Which graph represents acceleration (a) versus distance (x) variation of this system?
{{IMG1}}
- Option 1
- Option 2
- Option 3
- Option 4
### Related Formula
$$a = v \frac{dv}{dx}$$
### Core Logic
{{SOL_IMG1}}
Equation of $V$ vs $x$ from the provided graph is a straight line with a negative slope:
$V = C_1 - C_2 x$
(where $C_1, C_2 > 0$).
### Step 1: Calculate Acceleration
Differentiate $V$ with respect to $x$:
$$\frac{dV}{dx} = -C_2$$
Substitute into the acceleration formula:
$$a = (C_1 - C_2 x) \times (-C_2)$$
$$a = C_2^2 x - C_1 C_2$$
### Step 2: Analyze Graph Profile
The equation $a = C_2^2 x - C_1 C_2$ represents a straight line with a **positive slope** ($C_2^2$) and a **negative y-intercept** ($-C_1 C_2$).
{{SOL_IMG2}}
Therefore, the graph is a straight line intercepting the negative a-axis and moving positively.
### Pattern Recognition
A linear negative $v-x$ graph always yields an $a-x$ graph which is linear with a positive slope and negative intercept. Memorizing $a = v(dv/dx)$ maps graphical traits instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinematics
Q30 (2026)
Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from ground, at the instant when the first drop strikes the ground is ____ m.
$(g = 10 \mathrm{~m/s}^{2})$
- $2.5$
- $4.0$
- $4.2$
- $3.8$
### Related Formula
$$h = ut + \frac{1}{2}gt^2$$
### Core Logic
Since drops fall at regular intervals, find the total time of fall for the first drop and divide it by the number of intervals to get the time gap between successive drops.
{{SOL_IMG1}}
### Step 1: Total Time of Fall
Time taken by the first drop to reach the ground ($h = 5 \mathrm{~m}$):
$$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5}{10}} = 1 \mathrm{~sec}$$
### Step 2: Time Interval Between Drops
The first drop strikes the floor when the sixth drop begins to fall. There are exactly 5 intervals between the 1st and 6th drop.
Time between each drop $\Delta t = \frac{1 \mathrm{~sec}}{5} = 0.2 \mathrm{~sec}$.
### Step 3: Distance of the Fourth Drop
The 4th drop has been falling for 2 intervals (since drops 6, 5, 4 mean the 4th drop was released $2 \times 0.2 \mathrm{~s}$ after the 6th drop started... wait, no. The 4th drop was released before the 5th and 6th).
Time of fall for the $4^{\text{th}}$ drop is:
It was released 2 intervals after the 1st drop, so it has been falling for $1.0 - 2(0.2) = 0.6 \mathrm{~s}$? Wait, the PDF solution says:
"Time of fall for $4^{\text{th}}$ drop is $1 - 0.6 = 0.4 \mathrm{~sec}$"
Let's trace: 6th drop (0s fall), 5th drop (0.2s fall), 4th drop (0.4s fall).
Distance fallen by $4^{\text{th}}$ drop:
$$h' = \frac{1}{2} g t^2 = \frac{1}{2} \times 10 \times (0.4)^2 = 0.8 \mathrm{~m}$$
### Step 4: Height from Ground
Height from ground = Total height - Distance fallen
$$H = 5 - 0.8 = 4.2 \mathrm{~m}$$
### Pattern Recognition
In 'falling drops' problems, if N drops are in the air, there are (N-1) time intervals. The $k^{\text{th}}$ drop from the top has fallen for $(k-1)$ intervals.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Motion in a Straight Line
Q33 (2026)
A particle starts moving from time $t = 0$ and its coordinate is given as $x(t) = 4t^{3} - 3t$ .
A. The particle returns to its original position (origin) $0.866$ units later
B. The particle is 1 unit away from origin at its turning point.
C. Acceleration of the particle is non-negative.
D. The particle is 0.5 units away from origin at its turning point.
E. Particle never turns back as acceleration is non-negative.
Choose the correct answer from the options given below:
- $\text{A,C,D only}$
- $\text{A,B,C only}$
- $\text{C,E only}$
- $\text{A,C only}$
### Related Formula
$$v = \frac{dx}{dt}$$
$$a = \frac{dv}{dt}$$
### Core Logic
Given $x(t) = 4t^3 - 3t$.
The particle returns to origin when $x = 0$:
$$4t^3 - 3t = 0 \Rightarrow t(4t^2 - 3) = 0$$
$t = 0$ (start) and $t = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \approx 0.866 \text{ s}$. Thus, statement A is correct.
### Step 1: Velocity and Turning Point
Velocity is $v = \frac{dx}{dt} = 12t^2 - 3$.
At turning point, velocity becomes zero:
$$12t^2 - 3 = 0 \Rightarrow t^2 = \frac{1}{4} \Rightarrow t = \frac{1}{2} \text{ s}$$
(We take $t > 0$ since motion starts at $t=0$).
### Step 2: Position at Turning Point
Substitute $t = 1/2$ into $x(t)$:
$$x\left(\frac{1}{2}\right) = 4\left(\frac{1}{8}\right) - 3\left(\frac{1}{2}\right) = \frac{1}{2} - \frac{3}{2} = -1$$
The particle is $|-1| = 1$ unit away from origin. Statement B is correct, and D is incorrect.
### Step 3: Acceleration check
Acceleration $a = \frac{dv}{dt} = 24t$.
Since $t \geq 0$, $a \geq 0$, so acceleration is always non-negative. Statement C is correct.
Statement E is incorrect because the particle does turn back (at $t = 0.5$, velocity changes sign from negative to positive).
### Step 4: Final Conclusion
Statements A, B, and C are correct.
### Pattern Recognition
To analyze 1D motion, simply find roots of $x(t)=0$ (origin passes), $v(t)=0$ (turning points), and check the sign of $a(t)$ over the given domain $t \geq 0$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinematics
